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I recently learned about set exponentiation $B^{A}$, that is,

$B^{A} = \left \{ f \mid f:A\rightarrow B \right \}$

which is the set of all the functions which maps A to B. The number of function in a set exponentiation is given as $ \mid B \mid ^{\mid A \mid}$.

Let consider these sets S = {1,2,3} and T = {4,5}, then according to set exponentiation there are $2^3=8$ functions which map the above sets S to T.

Question 1: Can you please provide a simple example function which maps the above sets S and T? else can you provide another example for the same?

Question 2: How mathematicians proved the number of functions in set exponentiation is $ \mid B \mid ^{\mid A \mid}$ ?

JMoravitz
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Eka
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  • Just map elements to elements.. – Cameron L. Williams May 05 '21 at 14:59
  • As an aside, this is not explicitly "exponentiation"... The notation happens to resemble exponentiation and was likely chosen specifically for the convenience of having $|B^A|=|B|^{|A|}$, but it is built on more fundamental principles and different principles than the definitions used in arithmetic involving the number $e$. – JMoravitz May 05 '21 at 15:03
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    For finite sets it's simply a matter of noting that for each of the $|A|$ elements of of $A$ you have $|B|$ choices of elements of $B$ then the each element of $A$ may be mapped to. The eight functions of $S\to T$ are simply the $8$ choices that $1,2,3$ can be mapped to either $4$ or $5$. That is: 1: $1\to 4;2\to 4;3\to4$, 2: $1\to 4; 2\to 4; 3\to 5;$ 3: $1\to 4; 2\to 5;3\to 4$; 4: $1\to 4; 2\to 5; 3\to 5;$ 5: $1\to 5;2\to 4;3\to 4;$ 6: $1\to 5;2\to 4;3\to 5;$ 7:$1\to 5;2\to 5;3\to 4;$ 8: $1\to 5;2\to 5; 3\to 5$.... That's all. – fleablood May 05 '21 at 15:40
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    Suppose the question were "How many ways can you make a three character string using the the two characters $4$ and $5$?". The answer is for each position you have $2$ choices and you have three positions so you have three times you make the choice so the string is "${4,5},{4,5},{4,5}$ and there are $2\cdot 2\cdot 2 = 2^3$ ways to do this.... Well, a string of three characters is nothing more or less than a function between string position $\to$ character value. In the domain you have three possible positions and in the range two possible characters. That's $2^3$ possible functions. – fleablood May 05 '21 at 15:47
  • That is.... the eight functions are: $123\to 444; 123\to 445; 123\to 454; 123\to 455; 123\to 544;123\to 545; 123\to 554; 123\to 555$. – fleablood May 05 '21 at 15:50

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An example of a function $f:S \to T$ is given by $$f(1) = 4, \ f(2) = 5, \ f(3) = 4.$$ To see why $\mathrm{card} \ T^S = (\mathrm{card} \ T)^{\mathrm{card} \ S}$, consider how we could define a function $g: S \to T$. We could set $g(1) = 4 \text{ or } 5$. Similarly, there are two options for the value $g(2)$ and two for the value of $g(3)$. Therefore, the number of different functions one could define is $$2\cdot2\cdot2 = 2^3 = (\mathrm{card} \ T)^{\mathrm{card} \ S}.$$

Edit: Of course, this logic can be extended to larger sets $S$ and $T$. The link provided in the comment has a good answer for that.

Gary Moon
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